Haskell Diagrams: arrows with fixed orientation

Question

I need to draw arrows between two arbitrary "nodes". The arrow ends needs to enter or exit the nodes from one of the four cardinal directions: N, S, E, W.

data Dir = N | S | E | W
     deriving (Eq, Ord, Show)

cir, circles :: Diagram B
cir  = circle 0.3 # showOrigin # lw thick
circles = (cir # named "1") ||| strutX 3 ||| (cir # named "2")

ctrlPoint :: Dir -> V2 Double
ctrlPoint N = r2 (0, 1)
ctrlPoint S = r2 (0, -1)
ctrlPoint E = r2 (1, 0)
ctrlPoint W = r2 (-1, 0)

-- This function should specify an arrow shaft entering nodes from directions dir1 and dir2
shaft :: Dir -> Dir -> Trail V2 Double
shaft dir1 dir2 = trailFromSegments [bézier3 (controlPoint dir1) (controlPoint dir2) (r2 (3, 0))]

example = circles # connect' (with ... & arrowShaft .~ shaft N S ) "1" "2"

In the picture above, the arrow enters correctly from North in the first circle, and South in the second. However, if I setup the points vertically, everything is rotated:

circles = (cir # named "1") === strutY 3 === (cir # named "2")

This is not correct, because I wanted the arrow to enter from North and South, respectively. It seems the shaft of the arrow is rotated altogether... How to write my function shaft :: Dir -> Dir -> Trail V2 Double? Thanks

Answer 1

I found an answer using arrowFromLocatedTrail' instead:

-- control points for bézier curves
control :: Dir -> V2 Double
control N = r2 (0, 0.5)
control S = r2 (0, -0.5)
control E = r2 (0.5, 0)
control W = r2 (-0.5, 0)

-- shaft of arrows
shaft :: (P2 Double, Dir) -> (P2 Double, Dir) ->  Located (Trail V2 Double)
shaft (p, d) (p', d') = trailFromSegments [bézier3 (control d) ((p' .-. p) - (control d')) (p' .-. p)] `at` p

-- create a single arrow
mkArrow :: (P2 Double, Dir) -> (P2 Double, Dir) -> Diagram B
mkArrow a b = arrowFromLocatedTrail' (with & arrowHead .~ dart
                                    & lengths .~ veryLarge
                                    & shaftStyle %~ lw thick) (shaft a b)

This version performs the necessary transformations:

bézier3 (control d) ((p' .-. p) + (control d')) (p' .-. p)

Here is the signature ofbézier:

bézier3 :: v n -> v n -> v n -> Segment Closed v n 

It takes 3 vectors, named here V1, V2 and V3. bézier curve are by default not located in Diagrams, they just specify how to move.

So, to draw the bézier curve, we set:

V1 = control d
V2 = (p' .-. p) + (control d')
V3 = p' .-. p

The resulting bézier curve will located at p.